I am trying to simplify the following expression \begin{align} P\left(\bigcap_{i\in A}\{X_i\ge a\}\cap\bigcap_{i\in \overline A}\{X_i\le a\}\right) \end{align} where $A\subseteq\mathcal{A}$ and $\overline A = \mathcal{A}\setminus A$. All random variables $X_i$ are statistically independent.
My confusion arises from the complement sets. If the argument was just $\bigcap_{i\in A}\{X_i\ge a\}$ then I could write the probability as a product, but is the complement term interfering with this independence?
If $A$ and $B$ are independent, then also $A,\overline{B}$ and $\overline{A},B$ and $\overline{A},\overline{B}$. This also holds for a family of independent sets.
But be aware that $\{X_i \le a\}$ is not the complement of $\{X_i \ge a\}$.
But by definition the $X_i$ are independent iff all sets contained in the generated $\sigma$-algebras are as well.
So take here $\mathcal{A_i} = \sigma(X_i)$ and $B_i = \{X_i \ge a\}$ for $i\in A$ resp. $B_i = \{X_i \le a\}$ for $i\in\overline{A}$ then $B_i \in \mathcal{A}_i$ and the $\mathcal{A}_i$ are independent by assumption… hence the $B_i$ are.
And so you have $$\bigcap_{i\in A}\{X_i\ge a\}\cap\bigcap_{i\in \overline A}\{X_i\le a\} = \bigcap_{i \in \mathcal{A}} B_i$$ and the $B_i$ are independent.
Or in short: All sets above are independent and you can write it as a product…